Answer:
the answer is 5
Step-by-step explanation:
5.3x5=26.5
Answer:
Step-by-step explanation:
The easiest way to fill out the graph is to find the slope.
The slope is y = 2x + 1.
Insert the x values of the graph for x in the equation to find y.
y = 2(-6) + 1 = -12 + 1 = -11. (-6, -11).
y = 2(-2) + 1 = -4 + 1= -3. (-2, -3).
Then, for finding x, just insert y.
1 = 2x + 1. Subtract 1 from both sides to isolate x.
0 = 2x. x = 0. (0, 1).
y = 2(2) + 1 = 4 + 1 = 5. (2, 5)
3 = 2x + 1. Subtract 1 from both sides.
2 = 2x. x = 1. (1, 3).
The slope of this function is 2x. The y-intercept is 1. The x-intercept is -2.
Answer: x=-4
Step-by-step explanation: first you collect like terms 8x-7x=1-5 then we do 8x-7x which would equal 1x or just x next we do 1-5 which equals 4 so the solution to this equation is x=-4
Wow !
OK. The line-up on the bench has two "zones" ...
-- One zone, consisting of exactly two people, the teacher and the difficult student.
Their identities don't change, and their arrangement doesn't change.
-- The other zone, consisting of the other 9 students.
They can line up in any possible way.
How many ways can you line up 9 students ?
The first one can be any one of 9. For each of these . . .
The second one can be any one of the remaining 8. For each of these . . .
The third one can be any one of the remaining 7. For each of these . . .
The fourth one can be any one of the remaining 6. For each of these . . .
The fifth one can be any one of the remaining 5. For each of these . . .
The sixth one can be any one of the remaining 4. For each of these . . .
The seventh one can be any one of the remaining 3. For each of these . . .
The eighth one can be either of the remaining 2. For each of these . . .
The ninth one must be the only one remaining student.
The total number of possible line-ups is
(9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) = 9! = 362,880 .
But wait ! We're not done yet !
For each possible line-up, the teacher and the difficult student can sit
-- On the left end,
-- Between the 1st and 2nd students in the lineup,
-- Between the 2nd and 3rd students in the lineup,
-- Between the 3rd and 4th students in the lineup,
-- Between the 4th and 5th students in the lineup,
-- Between the 5th and 6th students in the lineup,
-- Between the 6th and 7th students in the lineup,
-- Between the 7th and 8th students in the lineup,
-- Between the 8th and 9th students in the lineup,
-- On the right end.
That's 10 different places to put the teacher and the difficult student,
in EACH possible line-up of the other 9 .
So the total total number of ways to do this is
(362,880) x (10) = 3,628,800 ways.
If they sit a different way at every game, the class can see a bunch of games
without duplicating their seating arrangement !
